Question

You measure 40 watermelons' weights, and find they have a mean weight of 67 ounces. Assume the population standard deviation is 11.4 ounces. Based on this, what is the maximal margin of error associated with a 99% confidence interval for the true population mean watermelon weight.

Answer 1

`35.36,44.64`

Explanation:

Sample size
`n=40`

Mean weight
`\=x =67`

Standard deviation
`\sigma=11.4`

Confidence Interval
`CI=0.99`

\alpha==0.01

Therefore

`Z_{(\alpha)/(2)}=Z_[0.005]`

From table

`Z_{(\alpha)/(2)}=2.576`

Generally, the equation for Margin of error is mathematically given by

`M.E=Z_{(\alpha)/(2)}*(\frac{\sigma}{sqrt{n}}`

`M.E=2.576*((11.4)/(√(40))`

`M.E=4.64`

Therefore Estimated mean is

`\=x-M.E<\mu <\=x +E`

`40-4.64<\mu< 40+4.64`

`35.36<\mu < 44.64`

`35.36,44.64`